Problem: Is ${49649}$ divisible by $3$ ?
Explanation: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {49649}= &&{4}\cdot10000+ \\&&{9}\cdot1000+ \\&&{6}\cdot100+ \\&&{4}\cdot10+ \\&&{9}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {49649}= &&{4}(9999+1)+ \\&&{9}(999+1)+ \\&&{6}(99+1)+ \\&&{4}(9+1)+ \\&&{9} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {49649}= &&\gray{4\cdot9999}+ \\&&\gray{9\cdot999}+ \\&&\gray{6\cdot99}+ \\&&\gray{4\cdot9}+ \\&& {4}+{9}+{6}+{4}+{9} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first four terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${49649}$ is divisible by $3$ if ${ 4}+{9}+{6}+{4}+{9}$ is divisible by $3$ Add the digits of ${49649}$ $ {4}+{9}+{6}+{4}+{9} = {32} $ If ${32}$ is divisible by $3$ , then ${49649}$ must also be divisible by $3$ ${32}$ is not divisible by $3$, therefore ${49649}$ must not be divisible by $3$.